In supplemental notes 23 and 26, we used a different coordinate system to hopefully simplify the integration that we were doing. So far we have used polar, cylindrical, and spherical coordinates, and these are special cases of a far more general substitution method.

figure 2

Suppose that a region G (See figure 1) in the uv-plane is transformed one-to-one into the region R (See figure 2) in the xy-plane by the equations x = g (u, v), y = h (u, v). R is the image of G under the transformation, and G is the pre-image of R. In fact, any function f (x, y) defined on R can be thought of as a function f (g (u, v), h (u, v)) defined on G.

The question that you should be asking yourself is this: How is the integral of f (x, y) over R related to the integral of f (g (u, v), h (u, v)) over G? If g, h, and f have continuous partial derivatives and J (u, v) is zero only at isolated points, if at all, then

where J (u, v) is the Jacobian of the coordinate transformation x = g (u, v), y = h (u, v). The Jacobian is defined to be

EXAMPLE 1:

Find the Jacobian for the transformation x = r cos q , y = r sin q used in polar coordinates.

SOLUTION:

EXAMPLE 2:

(A) Solve the system u = x - y, v = 2x + y for x and y in terms of u and v. Then find the value of the Jacobian, J (u, v) = ¶ (x, y)/¶ (u, v). (B) Find the image under the transformation u = x - y, v = 2x + y of the triangular region with vertices (0, 0), (1, 1), and (1, -2) in the xy-plane. Sketch the transformed region in the uv-plane.

SOLUTION: (A)

Notice that solving for x and y in terms of u and v can get kind of hairy, so here is a little trick that I learned in my calculus III course that I took. 

FACT: 

Isn't that little easier than the way we did it earlier?

Since I am given the vertices of the triangle in the original region, I must determine the equations of the lines that form the boundaries of this region. 

The lines that are the boundaries for the original region are x = 0, y = x, and y = -2x. Now to perform the transformation using u = x - y and v = 2x + y.

The boundaries for the new region are u = 0, v = 0, and u + v = 3.

Either region is nice, and the double integral over either region would not be hard to set up. The reason why I did this example is to show you how the method of substitution works before we start using it to set up integrals.

EXAMPLE 3: 

Use the transformation in example 2 to evaluate the integral

for the region R in the first quadrant bounded by the lines y = -2x + 4, y = -2x + 7, y = x - 2, and y = x + 1.

 SOLUTION:

Now to convert 2x ² - xy - y ².

Now to set up the integral and integrate.

EXAMPLE 4:

 Let R be the parallelogram in the xy-plane with boundaries x = -3, x = 0, y = x, and y = x + 1. Use the transformation u = 2x - 3y, v = -x + y to rewrite the integral

as an integral over an appropriate region G in the uv-plane. Then evaluate the uv-integral over G.

 First I will find the Jacobian for this transformation.

Now to transform the region.

Now to convert the integrand, and evaluate the new integral.

From supplemental notes 26, we used cylindrical and spherical coordinate substitutions. These are special cases of a substitution method that transforms the three-dimensional regions and changes the variables of the triple integral. The Jacobian for this transformation is

The trick that I have been using also works in the three-dimensional case.

EXAMPLE 5:

Find the Jacobian for the spherical coordinates x = r sin j cos q , y = r sinj sin q , z = r cos j .

EXAMPLE 6:

Find the Jacobian for the cylindrical coordinates x = r cos q , y = r sin q , z = z.

EXAMPLE 7:

Substitute u = xy, v = xy ³ to find the area of the first quadrant region bounded by the curves xy = 2, xy = 4, xy ³ = 3, and xy ³ = 6.